Skip to main content

Full text of "Text Book Of Mechanical Engineering"

See other formats

Weight of Fly Wheel.                        679

every radius the ordinates of both curves are added to form
the resulting curve cc. An average circle is struck, and shewn
dotted; and a clear conception of the more even turning move-
ment is then obtained.

Three cranks are set out at (#), Fig. 66j, and the like process
followed. The same letters are adopted throughout, and a more
regular turning movement results. The differences between the
c c curve and the dotted circle may seem little better than before,
but they form a much smaller percentage of the effort ordinate.

Calculation of Fly-wheel Weight required.—The
crank radius, Fig. 660, being i| feet, the circumference of the
crank circle is exactly n feet. In Fig. 662, let ad be 22 feet,
and let it be divided so that ae and hd are each 2| feet, and <?/
/£, and gA, are each 5 J- feet. On ef and g h set up ordinates
of crank effort on the up stroke, and on fg of that on the down
stroke, those on a e and • h d each representing half the down
stroke effort. Now take the mean of the ordinates on ef:
dividing the base into 10 parts, measuring at centre of each part,
adding the ordinates and dividing by 10 : the result is 29,500 Ibs.
The mean of the ordinates onfg is found similarly to be 25,000
Ibs. Adding and dividing by 2, gives 27,250, the mean effort
for the continuous diagram a d. Draw / k at this pressure
above ad.

Now the areas, A, c, &c., shew surplus work, while the crank
travels from I to #z, and from n to p respectively, while the areas
B, D, &c,, shew a work deficit between m n and p q. The fly
wheel must absorb the work A or c, and give it out again at B or D,
and thus tend to equalise the crank effort. The mean pressures
and distances traversed have been measured at A, B, c, and D, and
are shewn by work rectangles. The total surplus and the total
deficit of work per revolution ^ are each found to be 88,700 foot
pounds, and the greatest of the four work areas A, B, c, and D, is
D, or 49,560 foot pounds. This is the amount of energy
which the fly wheel must be able to deliver, such delivery
decreasing its velocity, while the absorption of energy will in like
manner increase it But the heavier the fly wheel, the less will be
the fluctuation of velocity; and the problem is to find the weight
of wheel which will absorb the surplus energy and re-deliver it